One of the theorems for a vector field to be conservative is that $$\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}$$ for $$F=\langle M,N\rangle.$$

To find the $$\int \limits_C\left(\frac{10y}{x^2+y^2}+x\right)dx-\left(\frac{10x}{x^2+y^2}+e^{5y}\right)dy$$ for any $C$ in the $x>0$ plane from $A(1,1)$ to $B(2,6)$ the vector field has to be conservative. But, the partial derivative are not equal to each other but they have a function $$f=10\tan^{-1}(x/y)+1/2x^2-1/5e^{5y}$$ for which the two components are satisfied. Is the vector field conservative/path independent? How would we find the line integral if it is not?

  • $\begingroup$ Why do you think that in order to calculate a line integral of a vector field this has to be conservative?? This is false, of course, otherwise line integrals would be very, very limited. $\endgroup$ – DonAntonio Dec 12 '13 at 13:55
  • $\begingroup$ @DonAntonio The question has two parts: find the potential function and solve the integral, 2) use two easy paths to find the line integral. This implies that the vector field is path independent--but it seems that it isn't. But, I was able to find a function that satisfies the two components of the vector field, so I am lost is it conservative or is it not? $\endgroup$ – John Dec 12 '13 at 14:11
  • $\begingroup$ The partial derivatives are equal to each other. $\endgroup$ – Rax Adaam Feb 18 '16 at 17:07

To verify a vector field is conservative or not, use:

$$\nabla \times F = 0$$

or say

$$\begin{vmatrix} \frac{\partial}{\partial x}& \frac{\partial}{\partial y} \\ M& N \\\end{vmatrix} = 0$$

In this case, after my calculation, it is indeed conservative.

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  • $\begingroup$ Indeed, I didn't continue with equation I just assumed that there not the same so they are not equal to each other but that is false. Its misleading to say that must be equal, its better to portray it as you do. $\endgroup$ – John Dec 12 '13 at 14:38
  • $\begingroup$ And thank you for the determinant--I was trying to find a way to memorize the 3D version of it and now I can :) $\endgroup$ – John Dec 12 '13 at 14:40
  • $\begingroup$ @John Yeah, this determinant is the general case luckily. $\endgroup$ – Ray Dec 12 '13 at 14:42

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